DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

116 SELECTED ANSWERS2.4.8 Only circles. By Exercise 2 such a curve will also have constant curvature, and byMeusnier’s formula, Proposition 2.5, the angle φ between N and n = x is constant.Differentiating x · N = cos φ = const yields τ(x · B) =0. Either τ =0,inwhich casethe curve is planar, or else x·B =0,inwhich case x = ±N, soτ = N ′ ·B = ±x ′ ·B =±T · B =0. (In the latter case, the curve is a great circle.)3.1.1 a. 2π sin u 03.1.3 a. 1, b.,c. 3.3.2.1 a. The semicircle centered at (2, 0) of radius √ 5; d(P, Q) =ln ( (3 + √ 5)/2 ) .3.2.4 Show first that for an arclength parametrization κ g = u ′ (s)/v(s)+θ ′ (s). Take a circleof the form α(t) = ( a + b cos t, b(sin t +1) ) and show υ(t) =‖α ′ (t)‖ =1/(sin t + 1).3.2.9 κ g = coth R3.3.4 We have κ n =II(e 1 , e 1 )=−de 3 (e 1 ) · e 1 = ω 13 (e 1 ). Since e 3 = sin θe 2 + cos θe 3 ,the calculations of Exercise 3 show that ω 13 = sin θω 12 + cos θω 13 , so ω 13 (e 1 ) =sin θω 12 (e 1 )=κ sin θ. Here θ is the angle between e 3 and e 3 ,sothis agrees with ourprevious result.3.3.7 We have ω 1 = bdu(and ω 2 = (a)+ b cos u) dv, so ω 12 = − sin udv and dω 12 =cos ucos u− cos udu∧ dv = −ω 1 ∧ ω 2 ,soK =b(a + b cos u)b(a + b cos u) .3.4.2 a. Taking ξ = f gives us ∫ 10 f(t)2 dt =0. Since f(t) 2 ≥ 0 for all t, iff(t 0 ) ≠0,wehavean interval [t 0 − δ, t 0 + δ] onwhich f(t) 2 ≥ f(t 0 ) 2 /2, and so ∫ 10 f(t)2 dt ≥ f(t 0 ) 2 δ>0.3.4.9 y = 1 2 cosh(2x)A.1.1A.1.2A.2.4Consider z = x − y. Then we know that z · v i =0,i =1, 2. Since {v 1 , v 2 } is a basisfor R 2 , there are scalars a and b so that z = av 1 + bv 2 . Then z · z = z · (av 1 + bv 2 )=a(z · v 1 )+b(z · v 2 )=0,soz = 0, asdesired.Hint: Take u =( √ a, √ b) and v =( √ b, √ a).Let v = ∫ ba f(t) dt. If v ≠ 0, we have ‖v‖2 = v · ∫ ba f(t) dt = ∫ bav · f(t) dt ≤∫ ba ‖v‖‖f(t)‖ dt = ‖v‖ ∫ ba‖f(t)‖ dt (using the Cauchy-Schwarz inequality u · v ≤‖u‖‖v‖), so ‖v‖ ≤ ∫ ba‖f(t)‖ dt, asneeded.